Find the expected value of a dice sum

Question

If fair dodecahedron is rolled until at least $k$($k$ is fixed between 2 and 12) is gotten, and $X$ is the sum of all numbers appeared until the last time, what is $E(X)$?

Answer 1

The probability that any roll is greater than or equal to $k$ is $$ \frac{13-k}{12} $$ so the expected number of rolls until a roll of $k$ or greater is $$ \frac{12}{13-k}. $$ All but the last one of these rolls is less than $k$, so the sum of those rolls has an expected value of $$ \left( \frac{12}{13-k} -1 \right) \frac{1+ (k-1)}{2}. $$ Add to this the expected value of the final roll $$ \frac{k+12}{2} $$ and so the expectation of the sum is $$ \left( \frac{12}{13-k} -1 \right) \frac{1+ (k-1)}{2} + \frac{k+12}{2} = \frac{78}{13-k}. $$

Answer 2

Let $n=12$ denote the number of faces. If the first roll is $i\geqslant k$, $X=i$. If the first roll is $i\lt k$, $X=i+X'$ where $X'$ is distributed like $X$. Hence, $$ \mathrm E(X)=\frac1n\sum_{i\geqslant k}i+\frac1n\sum_{i\lt k}\left(i+\mathrm E(X)\right)=\frac1n\sum_{i=1}^ni+\frac1n\mathrm E(X)\sum_{i=1}^{k-1}1, $$ that is, $$ n\mathrm E(X)=\frac{n(n+1)}2+(k-1)\mathrm E(X), $$ hence $$ \mathrm E(X)=\frac{n(n+1)}{2(n-k+1)}=\frac{78}{13-k}. $$